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I have some troubles solving the following problem:

Let $E$ be the elliptic curve $E:y^2+2y=x^3+x+9$ over $\mathbb{F}_{16}$. Compute the 2-torsion group $E[2]$, i.e. find all the points of order $2$ on $E$.

My idea is: if $2P=0$ then $P=-P$, so I have to find all the points that have this property. Usually if $P=(x,y)$ then $-P$ is defined as $-P=(x,-y-a_{1}x-a_{3})$. In our case $-P$ becomes $-P=(x,-y-2)$.

Can I say that since the characteristic of $\mathbb{F}_{16}$ is $2$ then $2y=0$ and $-y=y$ and then my equation becomes $E:y^2=x^3+x+9$ and $-P$ becomes $-P=(x,y)$? If this is correct, i'm right if I say that then obviously $P=-P$ for each point on the curve and so the 2-torsion group is simply $E(\mathbb{F}_{16})$?

Atlas
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    Something fishy here. The equation $y^2=x^3+x+1$ defines a singular curve (the partial derivative w.r.t. $y$ vanishes everywhere). Thus the group operation is suspect. In other words, if this part of a larger set of exercises working on a curve defined over the integers, it looks like it has bad reduction at $p=2$ (or was not put into a minimal form). – Jyrki Lahtonen Aug 25 '14 at 13:24
  • In characteristic two more often than not an elliptic curve is of the form $y^2+xy=x^3+ax+ b$. Then if $P=(x,y)$ we have $-P=(x,y+x)$. – Jyrki Lahtonen Aug 25 '14 at 13:26
  • Thank you Jyrki Lathonen! You're right, $y^2=x^3+x+1$ is not an elliptic curve because it isn't smooth, I didn't realize it. I even didn't think to check it! This exercise comes from an old exam,I quite don't understand the aim of it, but anyway...Thank you! – Atlas Aug 25 '14 at 15:03

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